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A circle is drawn with origin as the center and radius 10 units. Show that the point (–8, –6) is a point on the circle. Find out whether the point (9, –1) lies within the circle or outside the circle. Why? |
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Answer» The distance between the centre and the point (-8,-6) is = \(\sqrt{(-8)^2+(-6)^2}\) = \(\sqrt{100}\) = 10 It is equal to the radius. So the point (-8, -6) is a point on the circle. The distance between the centre, and the point (9,-1) is = \(\sqrt{9^2+(-1)^2}\) = \(\sqrt{81+1}\) = \(\sqrt{82}\) √82 is smaller than 10. So the point (9, –1) is a point inside the circle. Distance between (x1, y1) and (x3, y2) is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) Distance of (x, y) from (0, 0) is \(\sqrt{x^2+y^2}\) |
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